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(C) The orbital period $T$ of a satellite at a distance $r$ from the center of the Earth is given by Kepler's Third Law: $T = 2\pi \sqrt{\frac{r^3}{GM

Vedclass · Accepted Answer

${\left( {\frac{{GM{T^2}}}{{4{\pi ^2}}}} \right)^{1/3}} - R$

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(C) The orbital period $T$ of a satellite at a distance $r$ from the center of the Earth is given by Kepler's Third Law: $T = 2\pi \sqrt{\frac{r^3}{GM}}$.Here,$r = R + h$,where $R$ is the radius of the Earth and $h$ is the height of the satellite above the surface.Squaring both sides,we get: $T^2 = \frac{4\pi^2 r^3}{GM} = \frac{4\pi^2 (R+h)^3}{GM}$.Rearranging for $(R+h)^3$: $(R+h)^3 = \frac{GMT^2}{4\pi^2}$.Taking the cube root of both sides: $R+h = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$.Finally,solving for $h$: $h = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} - R$.

Given the radius of Earth $R$ and the length of a day $T$,the height of a geostationary satellite is: [$G$ = Gravitational Constant,$M$ = Mass of Earth]

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